If the lines 2x−3y=5 and 3x−4y=7 are two diameters of a circle of radius 7, then the equation of the circle is ?
Let (x0,y0) be the center of the circle
Intersection of any 2 diameter gives us the center.
x=5+3y2 substituting this in
3x–4y=7
35+3y2–4y=7
⟹y+1=0⟹y=−1
x0=5+3(−1)2=1
Equation of circle will be
(x–x0)2+(y–y0)2=r2
(x–1)2+(y+1)2=72
⟹x2+y2−2x+2y–47=0